Question

A wire of length $1\ \text{m}$ is moving at a speed of $2\ \text{m/s}$ perpendicular to a homogeneous magnetic field of $0.5\ \text{T}$. If the ends of the wire are joined to a resistance of $6\ \Omega$, the rate at which work is being done to keep the wire moving at that speed is

• A. $\frac{1}{3}\ \text{W}$
• B. $\frac{1}{6}\ \text{W}$
• C. $\frac{1}{12}\ \text{W}$
• D. $1\ \text{W}$

Hint:

To ensure accuracy, you can compute this problem using two completely independent paths: the electrical dissipation path ($P = e^2/R$) and the mechanical mechanical path ($P = Fv$). Since both methods yield exactly $\frac{1}{6}\ \text{W}$, you can be fully confident in your answer during a test!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Verify the power dissipated in a system by cross-checking through two independent physical approaches.

Step 2: Key Formula or Approach:
Approach 1 (Electrical): Power dissipated P = e²/R, where e is the induced emf and R is the circuit resistance. Approach 2 (Mechanical): Power delivered P = F·v, where F is the applied force and v is the velocity.

Step 3: Detailed Explanation:
Calculating via the electrical dissipation path gives P = e²/R = 1/6 W. Computing through the mechanical work rate path yields P = Fv = 1/6 W as well. Both completely independent methods converge on the exact same numerical value, providing strong confirmation. This dual-path verification ensures accuracy even under exam pressure.

Step 4: Final Answer:
Both methods confirm the power as 1/6 W.